In triangle $ABC$, $BC = 4$, $AC = 3 \sqrt{2}$, and $\angle C = 45^\circ$.  Altitudes $AD$, $BE$, and $CF$ intersect at the orthocenter $H$.  Find $AH:HD$.
Explanation: Since $\angle C = 45^\circ$, triangle $ACD$ is a $45^\circ$-$45^\circ$-$90^\circ$ triangle, which means $AD = CD = AC/\sqrt{2} = 3$.  Then $BD = BC - CD = 4 - 3 = 1$.

[asy]
unitsize(1 cm);

pair A, B, C, D, E, F, H;

A = (1,3);
B = (0,0);
C = (4,0);
D = (A + reflect(B,C)*(A))/2;
E = (B + reflect(C,A)*(B))/2;
F = (C + reflect(A,B)*(C))/2;
H = extension(B,E,C,F);

draw(A--B--C--cycle);
draw(A--D);
draw(B--E);
draw(C--F);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
label("$H$", H, SE);
[/asy]

Also, $\angle EBC = 90^\circ - \angle BCE = 45^\circ$, so triangle $BHD$ is a $45^\circ$-$45^\circ$-$90^\circ$ triangle.  Hence, $HD = BD = 1$.  Then $AH = AD - HD = 3 - 1 = 2$, so $AH:HD = \boxed{2}$.